Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $k \neq 0$. $x = \dfrac{k - 6}{k^2 - 8k + 16} \div \dfrac{k - 6}{-7k + 28} $
Solution: Dividing by an expression is the same as multiplying by its inverse. $x = \dfrac{k - 6}{k^2 - 8k + 16} \times \dfrac{-7k + 28}{k - 6} $ First factor the quadratic. $x = \dfrac{k - 6}{(k - 4)(k - 4)} \times \dfrac{-7k + 28}{k - 6} $ Then factor out any other terms. $x = \dfrac{k - 6}{(k - 4)(k - 4)} \times \dfrac{-7(k - 4)}{k - 6} $ Then multiply the two numerators and multiply the two denominators. $x = \dfrac{ (k - 6) \times -7(k - 4) } { (k - 4)(k - 4) \times (k - 6) } $ $x = \dfrac{ -7(k - 6)(k - 4)}{ (k - 4)(k - 4)(k - 6)} $ Notice that $(k - 6)$ and $(k - 4)$ appear in both the numerator and denominator so we can cancel them. $x = \dfrac{ -7(k - 6)\cancel{(k - 4)}}{ \cancel{(k - 4)}(k - 4)(k - 6)} $ We are dividing by $k - 4$ , so $k - 4 \neq 0$ Therefore, $k \neq 4$ $x = \dfrac{ -7\cancel{(k - 6)}\cancel{(k - 4)}}{ \cancel{(k - 4)}(k - 4)\cancel{(k - 6)}} $ We are dividing by $k - 6$ , so $k - 6 \neq 0$ Therefore, $k \neq 6$ $x = \dfrac{-7}{k - 4} ; \space k \neq 4 ; \space k \neq 6 $